From different sets of data of t1/2 at different initial concentration say 'a' for a given reaction, the product [ t1/2 × a] is found to be constant. The order of reaction is

Question

From different sets of data of t1/2 at different initial concentration say 'a' for a given reaction, the product [ t1/2 × a] is found to be constant. The order of reaction is (A)  2 (B)  1 (C) zero (D)  3

Tardigrade Admin · Accepted Answer

For zero order, t1/2= (a/2k) So, t1/2 × a = (a2/2k) is not a constant. For 1st order, t1/2 is constant so t1/2 × a is not constant For 2nd order, t1/2 = (I/a× k) t 1/2× a = k is constant.

Q. From different sets of data of $t_{1/2}$ at different initial concentration say $^{'} a^{'}$ for a given reaction, the product $[t_{1/2} \times a]$ is found to be constant. The order of reaction is

$2$

$1$

zero

$3$

For zero order, $t_{1/2} = \frac{a}{2 k}$
So, $t_{1/2} \times a = \frac{a ^{2}}{2 k}$ is not a constant.
For $1^{s t}$ order, $t_{1/2}$ is constant so $t_{1/2} \times a$ is not constant
For $2^{n d}$ order, $t_{1/2} = \frac{I}{a \times k}$
$t_{1/2} \times a = k$ is constant.

Q. From different sets of data of t1/2​ at different initial concentration say ′a′ for a given reaction, the product [t1/2​×a] is found to be constant. The order of reaction is

2

1

zero

3

For zero order, t1/2​=2ka​ So, t1/2​×a=2ka2​ is not a constant. For 1st order, t1/2​ is constant so t1/2​×a is not constant For 2nd order, t1/2​=a×kI​ t1/2​×a=k is constant.

Q. From different sets of data of $t_{1/2}$ at different initial concentration say $^{'} a^{'}$ for a given reaction, the product $[t_{1/2} \times a]$ is found to be constant. The order of reaction is

$2$

$1$

$3$

For zero order, $t_{1/2} = \frac{a}{2 k}$
So, $t_{1/2} \times a = \frac{a ^{2}}{2 k}$ is not a constant.
For $1^{s t}$ order, $t_{1/2}$ is constant so $t_{1/2} \times a$ is not constant
For $2^{n d}$ order, $t_{1/2} = \frac{I}{a \times k}$
$t_{1/2} \times a = k$ is constant.